what volume of a 0.100 M solution of NaHCO3 contains 0.350 g of NaHCO3?

Answer 1

The molar mass Mr of NaHCO3 is Mr = 84 g/mol. So the 0.350 g of NaHCO3 are:

n = m/Mr, n = 0.35/84 mol or n = 4.17*10^(-3) mol.

Now, n = C*V, where C = 0.1 M the molarity and V the volume of the solution, so:

V = n/C, V = 4.17*10^(-3)/0.1, V = 0.0417 L or 41.7 mL